Question

.If the graph of equation 2x + ky = 10k, intersect x-axis at point (2, 0) then find value of k.​

Answer 1

Answer:

hiii

your answer is here !

Step-by-step explanation:

Here,

2x + ky = 10k ...(i)

2x + ky = 10k ...(i)At point (2, 0), equation (i) becomes:

=> 2(2) + k(0) = 10k

⟹ 4 = 10k

⟹ 4 = 10k⟹ k = 10/4 = 0.4

follow me !

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1cm}\begin{picture}(6,8)\thicklines\put(1, 2.8){\line(2,0){5}}\put(3.5,1){\line(0,2){4}}\put(3.8,1){\line(8,3){1.4}}\put(1,2.8){\vector(-2,0){0.5}}\put(6,2.8){\vector(2,0){0.5}}\put(3.97,2.7){\line(0,2){0.2}}\put(4.4,2.7){\line(0,2){0.2}}\put(3.5,1){\vector(0,-2){0.5}}\put(3.5,5){\vector(0,2){0.5}}\put(4.4,2.8){\circle*{0.1}}\put(4.4,2.4){\sf{(2,0)}}\put(4,1.2){\sf{2x + ky = 10k}}\end{picture}$

$\rule{160}{1}$

If the Graph Intersect at x - axis at point (2, 0), then value of x = 2 and, y = 0.

$\dashrightarrow\tt\:\:2x+ky=10k\\\\{\scriptsize\qquad\bf{\dag}\:\:\textsf{Putting the value of x and y.}}\\\\\dashrightarrow\tt\:\:2(2)+k(0)=10k\\\\\\\dashrightarrow\tt\:\:4+0=10k\\\\\\\dashrightarrow\tt\:\:4=10k\\\\\\\dashrightarrow\tt\:\:\dfrac{4}{10}=k\\\\\\\dashrightarrow\:\:\underline{\boxed{\textsf{\textbf{k = 0.4}}}}$